mod of a cap - b cap = ¡Ì2 then calculate the value of mod of a cap + ¡Ì3b acp

To solve the given equation, we have:

|a| - |b| = √2

First, let's consider the case when a ≥ 0 and b ≥ 0:

In this case, the equation becomes:

a - b = √2

Next, let's consider the case when a ≥ 0 and b < 0:

In this case, the equation becomes:

a - (-b) = √2 (since the magnitude of a negative number is equal to its positive counterpart)

a + b = √2

Now, let's consider the case when a < 0 and b ≥ 0:

In this case, the equation becomes:

(-a) - b = √2 (since the magnitude of a negative number is equal to its positive counterpart)

-a - b = √2

Finally, let's consider the case when a < 0 and b < 0:

In this case, the equation becomes:

(-a) - (-b) = √2

-b + a = √2

Now, the value of |a| + √3|b| can be calculated in each of the above cases.

- When a ≥ 0 and b ≥ 0:

|a| + √3|b| = a + √3b

- When a ≥ 0 and b < 0:

|a| + √3|b| = a + √3(-b) = a - √3b

- When a < 0 and b ≥ 0:

|a| + √3|b| = (-a) + √3b = -a + √3b

- When a < 0 and b < 0:

|a| + √3|b| = (-a) + √3(-b) = -a - √3b

Therefore, the value of |a| + √3|b| depends on the cases discussed above.

To calculate the value of |â + √3b̂|, we need to find the values of â and b̂ first. Given the equation:

|â - b̂| = √2

We square both sides of the equation to eliminate the square root:

(â - b̂)² = (√2)²
â² - 2âb̂ + b̂² = 2

Now, let's multiply the equation by 3:

3(â² - 2âb̂ + b̂²) = 3(2)
3â² - 6âb̂ + 3b̂² = 6

Now, we are going to work with the expression |â + √3b̂|. We square both sides of the equation:

(|â + √3b̂|)² = (3â² - 6âb̂ + 3b̂²)

Since the square of any value is always positive, we can remove the absolute value sign:

(â + √3b̂)² = 3â² - 6âb̂ + 3b̂²

Now, let's substitute the value of the right side of the equation with the equivalent expression from the previous equation:

(â + √3b̂)² = 6

Taking the square root of both sides, we have:

√(â + √3b̂)² = √6
(â + √3b̂) = ±√6

Therefore, the value of |â + √3b̂| is ±√6.